# Chemical Thermodynamics | General Chemistry 3

Chemical thermodynamics is studied in this chapter: spontaneity of reactions, first and second law of thermodynamics, entropy, Gibbs energy and standard Gibbs energy change, relationship between Gibbs energy and reaction quotient, Gibbs energies of formation, van’t Hoff equation

## Spontaneity of Reactions

Spontaneous process: a process that takes place without the input of energy from an external source
It generally leads to a decrease in the energy of the system ⇒ a spontaneous process is very often exothermic. But it's not always the case

2 H2 (g) + O2 (g) → 2 H2O (l) is a spontaneous process
ΔH0rxn = - 571.6 kJ.mol-1 < 0 ⇒ exothermic reaction

H2O (s) → H2O (l) is a spontaneous process at T > 0°C
ΔH0fus = +6.0 kJ.mol-1 > 0 ⇒ not exothermic reaction

## Law of Thermodynamics

First Law of Thermodynamics:

The energy of the universe is constant ⇒ energy can neither be created nor destroyed
ΔUuniv = ΔUsys + ΔUsurr = 0
⇒ ΔUsys = - ΔUsurr = q + w

Second Law of Thermodynamics:

The total entropy of an isolated system can never decrease over time and is constant if all processes are reversible
⇒ the entropy change is never negative: ΔSsystem + ΔSsurrounding $\ge$ 0
(ΔSsystem + ΔSsurrounding = 0 for an equilibrium)

Entropy S (in J.K-1): measure of the amount of disorder in a system (ex: expansion of a gas into a vacuum)

ΔSsys $\ge$ $\frac{{\mathrm{q}}_{\mathrm{sys}}}{{\mathrm{T}}_{\mathrm{sys}}}$

ΔSsys = entropy change (in J.K-1)
qsys = heat (in J)
Tsys = temperature (in K)

## Entropy change

Entropy change on fusion:

ΔSfus = $\frac{∆{\mathrm{H}}_{\mathrm{fus}}}{{\mathrm{T}}_{\mathrm{m}}}$

ΔSfus = molar entropy of fusion (in J.K-1.mol-1)
ΔHfus = molar enthalpy of fusion (in J.mol-1)
Tm = melting point (in K)

Entropy change on vaporization:

ΔSvap = $\frac{∆{\mathrm{H}}_{\mathrm{vap}}}{{\mathrm{T}}_{\mathrm{b}}}$

ΔSvap = molar entropy of vaporization (in J.K-1.mol-1)
ΔHvap = molar enthalpy of vaporization (in J.mol-1)
Tb = boiling point (in K)

## Entropy Changes for Reactions

Standard entropy change ΔS0rxn for a chemical reaction (in J.K-1.mol-1):

ΔS0rxn = αi S[products] - αi S[reactants]

αi = stoichiometric coefficient
S0 = standard entropy (in J.K-1.mol-1)

a A + b B → c C
ΔS0rxn = c S[C] – a S[A] – b S[B]

N(g) + 3 H2 (g) $⇌$ 2 NH3 (g)
ΔS0rxn = 2 S[NH3] – S[N2] – 3 S[H2]

## Gibbs Energy Change and Spontaneity of Reactions

Gibbs energy change ΔGrxn (in J.mol-1):

ΔGrxn = ΔHrxn - TΔSrxn

ΔHrxn = enthalpy change (in J.mol-1)
ΔSrxn = entropy change (in J.K-1.mol-1)
T = temperature (in K)

Gibbs criteria for reaction spontaneity:

• If ΔGrxn < 0: reaction is spontaneous and additional products can form
• If ΔGrxn > 0: reaction is not spontaneous ⇒ Input of energy from an external source is necessary to form additional products
• If ΔGrxn = 0: reaction is at equilibrium

## Gibbs Energy Change and Reaction Quotient

ΔGrxn = RT ln $\left(\frac{\mathrm{Q}}{\mathrm{K}}\right)$

R = ideal gas constant = 8.314 J.mol-1.K-1
T = temperature (in K)
Q = reaction quotient
K = equilibrium constant

If Q > K  $⇔$  ΔGrxn > 0: reaction proceeds spontaneously from right to left
If Q < K   $⇔$   ΔGrxn < 0: reaction proceeds spontaneously from left to right
If Q = K   $⇔$   ΔGrxn = 0: reaction is at equilibrium
A reversible chemical reaction has reached equilibrium when the Gibbs free energy Grxn reaches a minimum

## Relationship between ΔG and ΔG°

Standard Gibbs energy change ΔG0rxn for a reaction (in J.mol-1):

ΔG0rxn = - RT ln K

R = ideal gas constant = 8.314 J.mol-1.K-1
T = temperature (in K)
K = equilibrium constant

ΔGrxn = RT ln $\left(\frac{\mathrm{Q}}{\mathrm{K}}\right)$ = RT ln Q – RT ln K:

ΔGrxn = ΔG0rxn + RT ln Q

R = ideal gas constant = 8.314 J.mol-1.K-1
T = temperature (in K)
Q = reaction quotient

## Gibbs Energies of Formation

Standard Gibbs energy ΔG0rxn for a chemical reaction (in J.mol-1):

ΔG0rxn = αi ΔG0[products] - αi ΔG0[reactants]

αi = stoichiometric coefficient
ΔG0f = standard molar Gibbs energy of formation
(in J.mol-1)

a A + b B → c C
ΔG0rxn = c ΔG0[C] – a ΔG0[A] – b ΔG0[B]

N2 (g) + 3 H2 (g) $⇌$ 2 NH3 (g)
ΔG0rxn = 2 ΔG0[NH3] – ΔG0[N2] – 3 ΔG0[H2]

## Van’t Hoff and Clapeyron-Clausius Equations

ΔG0rxn = ΔH0rxn – TΔS0rxn and ΔG0rxn = - RT ln K

⇒ ln K = - $\frac{∆{{\mathrm{H}}^{0}}_{\mathrm{rxn}}}{\mathrm{RT}}$ + $\frac{∆{{\mathrm{S}}^{0}}_{\mathrm{rxn}}}{\mathrm{R}}$

⇒ a plot of ln K versus $\frac{1}{\mathrm{T}}$ is linear with slope - $\frac{∆{{\mathrm{H}}^{0}}_{\mathrm{rxn}}}{\mathrm{R}}$ and intercept $\frac{∆{{\mathrm{S}}^{0}}_{\mathrm{rxn}}}{\mathrm{R}}$

van’t Hoff Equation:

ln $\frac{{\mathrm{K}}_{1}}{{\mathrm{K}}_{2}}$ = $\frac{∆{{\mathrm{H}}^{0}}_{\mathrm{rxn}}}{\mathrm{R}}$ x  = $\frac{∆{{\mathrm{H}}^{0}}_{\mathrm{rxn}}}{\mathrm{R}}$ x

Clapeyron-Clausius Equation:

ln $\frac{{\mathrm{P}}_{2}}{{\mathrm{P}}_{1}}$= $\frac{∆{{\mathrm{H}}^{0}}_{\mathrm{rxn}}}{\mathrm{R}}$ x  = $\frac{∆{{\mathrm{H}}^{0}}_{\mathrm{rxn}}}{\mathrm{R}}$ x